If $f:\mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x)=x$ has no real solution, then show that $f(f(x))=x$ has no real solution either.
Is the proof trivial as it seems or does it need an analytical approach?
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asked Jun 4, 2016 at 11:14
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1$\begingroup$ Do you want a hint or a full solution? My hint is look at $f(x)-x$. $\endgroup$– EmreJun 4, 2016 at 11:16
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$\begingroup$ @Emre I need more than that, thanks. $\endgroup$– StubbornAtomJun 4, 2016 at 11:17
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1 Answer
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As $f$ is continuous, $g(x)=f(x)-x$ is also continuous. We know that $g(x)=0$ has no root. Thus, either $g(x)>0$ for all $x\in\mathbb{R}$ or $g(x)<0$ for all $x\in\mathbb{R}$. In the first case, we have $f(x)>x$ for all $x\in\mathbb{R}$. So, $$f(f(x))>f(x)>x$$ In the second case, we have $f(x)<x$ for all $x\in\mathbb{R}$. So, $$f(f(x))<f(x)<x$$
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$\begingroup$ I am having confusion. In your argument are you not assuming that the function is monotonically increasing when you say $f(x)>x \implies f(f(x))>f(x)$? $\endgroup$ Sep 13, 2019 at 15:59
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1$\begingroup$ No, I am replacing $x$ with $f(x)$ in the statement. $\endgroup$– EmreMar 30, 2020 at 2:53